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若$a_{1},a_{2},\dots,a_N\ge0,\;p>1$，则
$$\sum _{n=1}^N\left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{p}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^Na_{n}^{p}.$$
证明：见Wikipedia.$\quad\Box$
令$N\to\infty$得到：
若$a_{1},a_{2},\dots\ge0,\;p>1$，则
$$\sum _{n=1}^{\infty }\left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{p}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{\infty }a_{n}^{p}.$$
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